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Lời giải:
\(\frac{x^2+y^2}{xy}=\frac{10}{3}\Rightarrow 3(x^2+y^2)=10xy\)
\(\Leftrightarrow 3x^2-10xy+3y^2=0\)
Đặt \(x=ty\) thì \(3(ty)^2-10ty.y+3y^2=0\)
\(\Leftrightarrow y^2(3t^2-10t+3)=0\)
\(\Rightarrow 3t^2-10t+3=0\) (do $y\neq 0$)
\(\Leftrightarrow (t-3)(3t-1)=0\Rightarrow \left[\begin{matrix} t=3\\ t=\frac{1}{3}\end{matrix}\right.\)
\(B=\frac{x-y}{x+y}=\frac{ty-y}{ty+y}=\frac{t-1}{t+1}=\left[\begin{matrix} \frac{1}{2}\\ \frac{-1}{2}\end{matrix}\right.\)
Vậy..........
Lời giải:
Ta có \(\frac{x^2+y^2}{xy}=\frac{25}{12}\)
\(\Leftrightarrow 12(x^2+y^2)-25xy=0\)
\(\Leftrightarrow (3x-4y)(4x-3y)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-4y=0\\4x-3y=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4y}{3}\left(1\right)\\x=\dfrac{3y}{4}\left(2\right)\end{matrix}\right.\)
Với (1):
\(A=\frac{x-y}{x+y}=\frac{\frac{4}{3}y-y}{\frac{4}{3}y+y}=\frac{\frac{1}{3}y}{\frac{7}{3}y}=\frac{1}{7}\)
Với (2)
\(A=\frac{x-y}{x+y}=\frac{\frac{3}{4}y-y}{\frac{3}{4}y+y}=\frac{\frac{-1}{4}y}{\frac{7}{4}y}=\frac{-1}{7}\)
Vậy
\(A=\pm \frac{1}{7}\)
a/ \(B=\left(\dfrac{x^2}{y}-\dfrac{y^2}{x}\right)\left(\dfrac{x+y}{x^2+xy+y^2}+\dfrac{1}{x-y}\right)\)
\(=\dfrac{x^3-y^3}{xy}\cdot\dfrac{\left(x+y\right)\left(x-y\right)+x^2+xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{x^3-y^3}{xy}\cdot\dfrac{x^2-y^2+x^2+xy+y^2}{x^3-y^3}\)
\(=\dfrac{2x^2+xy}{xy}=\dfrac{x\left(2x+y\right)}{xy}=\dfrac{2x+y}{y}\)
b/ Khi x = -1/2 và y = 3 ta có:
\(B=\dfrac{2\cdot\left(-\dfrac{1}{2}\right)+3}{3}=\dfrac{-1+3}{3}=\dfrac{2}{3}\)
\(\dfrac{x^2+y^2}{xy}=\dfrac{10}{3}\Rightarrow\dfrac{x^2+y^2}{10}=\dfrac{xy}{3}\)
Đặt \(\dfrac{x^2+y^2}{10}=\dfrac{xy}{3}=k\) (k > 0)
\(\Rightarrow\left\{{}\begin{matrix}x^2+y^2=10k\\xy=3k\end{matrix}\right.\)
\(\Rightarrow x^2+y^2+2xy=10k+2.3k=16k\)
\(\Leftrightarrow\left(x+y\right)^2=16k\Rightarrow x+y=4\sqrt{k}\)
\(\Rightarrow x^2+y^2-2xy=10k-2.3k=4k\)
\(\Leftrightarrow\left(x-y\right)^2=4k\Rightarrow x-y=2\sqrt{k}\)
Ta có \(M=\dfrac{x-y}{x+y}=\dfrac{2\sqrt{k}}{4\sqrt{k}}=\dfrac{1}{2}\)
Theo bài ra , ta có :
\(P=\left(\dfrac{x^2}{x^2-y^2}+\dfrac{y}{x-y}\right):\dfrac{x^3-y^3}{x^5-x^4y-xy^4+y^5}\)ĐKXĐ \(x\ne\pm y\)
\(\Leftrightarrow P=\left(\dfrac{x^2}{\left(x-y\right)\left(x+y\right)}+\dfrac{y\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}\right):\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{x^4\left(x-y\right)-y^4\left(x-y\right)}\)
\(\Leftrightarrow P=\left(\dfrac{x^2+xy+y^2}{\left(x-y\right)\left(x+y\right)}\right):\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(x-y\right)\left(x^4-y^4\right)}\)
\(\Leftrightarrow P=\dfrac{x^2+xy+y^2}{\left(x-y\right)\left(x+y\right)}\times\dfrac{\left(x-y\right)\left(x^4-y^4\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(\Leftrightarrow P=\dfrac{x^4-y^4}{\left(x-y\right)\left(x+y\right)}\)\(\Leftrightarrow P=\dfrac{\left(x^2\right)^2-\left(y^2\right)^2}{\left(x-y\right)\left(x+y\right)}=\dfrac{\left(x^2-y^2\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x+y\right)}=x^2+y^2\)(1)
Ta có : \(x+y=5\Rightarrow\left(x+y\right)^2=25\Rightarrow x^2+y^2=25-2xy=25--1=26\)(Vì xy = -1/2)
Thay x2 + y2 = 26 vào (1) ta đk : P = 26
Vậy P = 26 khi x + y = 5 và xy = -1/2
\(P=\left(\dfrac{x^2+y\left(x+y\right)}{\left(x^2-y^2\right)}\right).\left(\dfrac{x^4\left(x-y\right)-y^4\left(x-y\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\right)\\ \)
\(P=\left(\dfrac{x^2+xy+y^2}{\left(x^2-y^2\right)}\right).\dfrac{\left(x^2-y^2\right)\left(x^2+y^2\right)}{\left(x^2+xy+y^2\right)}\)
\(P=x^2+y^2=\left(x+y\right)^2-2xy=25-2\left(-\dfrac{1}{2}\right)=26\)
A=(xy2+xy−x−yx2+xy)(xy2+xy−x−yx2+xy) : (y2x3−xy2+1x+y):xy
A=( \(\dfrac{x}{y\left(x+y\right)}\) - \(\dfrac{x-y}{x\left(x+y\right)}\)) : (\(\dfrac{y^2}{x\left(x-y\right)\left(x+y\right)}\)+\(\dfrac{1}{x+y}\)) : \(\dfrac{x}{y}\)
A=\(\dfrac{x^2-y\left(x-y\right)}{xy\left(x+y\right)}\) : \(\dfrac{y^2+x\left(x-y\right)}{x\left(x-y\right)\left(x+y\right)}\) : \(\dfrac{x}{y}\)
A = \(\dfrac{x^2-xy+y^2}{xy\left(x+y\right)}\) : \(\dfrac{y^2-xy+x^2}{x\left(x-y\right)\left(x+y\right)}\):\(\dfrac{x}{y}\)
A = \(\dfrac{x^2-xy+y^2}{xy\left(x+y\right)}\). \(\dfrac{x\left(x-y\right)\left(x+y\right)}{x^2-xy+y^2}\):\(\dfrac{x}{y}\)
A = \(\dfrac{x-y}{y}\) : \(\dfrac{x}{y}\)
A = \(\dfrac{x-y}{x}\)
A= 1 - \(\dfrac{y}{x}\)>1
=> y/x <0
=> xy<0 , x+y khác 0
\(\dfrac{x^2+y^2}{xy}=\dfrac{10}{3}\Leftrightarrow x^2+y^2=\dfrac{10}{3}xy\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+2xy=\dfrac{10}{3}xy+2xy\\x^2+y^2-2xy=\dfrac{10}{3}xy-2xy\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=\dfrac{16}{3}xy\\\left(x-y\right)^2=\dfrac{4}{3}xy\end{matrix}\right.\)
Do \(0< x< y\Rightarrow\left\{{}\begin{matrix}x-y>0\\x+y>0\end{matrix}\right.\) \(\Rightarrow B>0\)
\(B^2=\dfrac{\left(x-y\right)^2}{\left(x+y\right)^2}=\dfrac{\dfrac{4}{3}xy}{\dfrac{16}{3}xy}=\dfrac{1}{4}\Rightarrow B=\dfrac{1}{2}\)